Find the Maclaurin series for the following function

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The discussion focuses on finding the Maclaurin series for the function f(x) = ln(1-x^3) / x^2. The initial attempt yielded a series of [(-1)^(2n-1) (x)^(n)] / (n), but the professor provided a different result: [(-1)^(2n-1) (x)^(3n-2)] / (n). Participants suggest breaking down the steps, starting with the Maclaurin series for ln(1-x) and then adapting it for ln(1-x^3). Clarification on the derivation process is sought to understand the discrepancy in results. The conversation emphasizes the importance of correctly applying series expansions in calculus.
jorgegalvan93
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Homework Statement



f(x) =ln (1-x^3) / (x^2)

Homework Equations



Using the maclaurin series ln (1 +x) = Ʃ (-1)^(n-1) (x^n)/(n)

The Attempt at a Solution



the maclaurin series for the function i get is [(-1)^(2n-1) (x)^(n)] / (n)
however, the answer according to my prof is [(-1)^(2n-1) (x)^(3n-2)] / (n)
How?
 

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jorgegalvan93 said:

Homework Statement



f(x) =ln (1-x^3) / (x^2)

Homework Equations



Using the maclaurin series ln (1 +x) = Ʃ (-1)^(n-1) (x^n)/(n)

The Attempt at a Solution



the maclaurin series for the function i get is [(-1)^(2n-1) (x)^(n)] / (n)
however, the answer according to my prof is [(-1)^(2n-1) (x)^(3n-2)] / (n)
How?
How about showing your steps ?

What is the Maclaurin series for ln(1-x) ?

Then what is the Maclaurin series for ln(1-x3) ?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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